模板合集

矩阵乘法以及求逆(实数)

struct matrix{
    int n,m;
    double a[N][N];
    matrix(int nn,int mm):n(nn),m(mm){
        memset(a,0,sizeof(a));
    }
    matrix operator *(const matrix &x) const{
        matrix b(n,x.m);
        for(int k=1;k<=n;++k)
            for(int i=1;i<=n;++i)
                for(int j=1;j<=n;++j)
                    b.a[i][j]+=a[i][k]*x.a[k][j];
        return b;
    }
    matrix inverse(const matrix &x){
        assert(x.n==x.m);
        int n=x.n;
        double p[N][2*N];
        memset(p,0,sizeof(p));
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j){
                p[i][j]=x.a[i][j];
                p[i][j+n]=(i==j);
            }
        for(int i=1;i<=n;++i){
            int z=i;
            for(int j=i;j<=n;++j)
                if(fabs(p[z][i])<fabs(p[j][i]))
                    z=j;
            swap(p[i],p[z]);
            for(int j=2*n;j>=i;--j)
                p[i][j]/=p[i][i];
            for(int j=i+1;j<=n;++j)
                for(int k=2*n;k>=i;--k)
                    p[j][k]-=p[i][k]*p[j][i];
        }
        matrix ans(n,n);
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j)
                ans.a[i][j]=p[i][j+n];
        return ans;
    }
};

后缀自动机

int ch[N*2][26],pre[N*2],len[N*2],last=1,k=1;
void Append(int c)
{
	int p=last;last=++k;len[k]=len[p]+1;
	for(;!ch[p][c]&&p;p=pre[p])
		ch[p][c]=k;
	if(!p){pre[k]=1;return;}
	int np=ch[p][c];
	if(len[np]==len[p]+1){pre[k]=np;return;}
	int q=++k;
	memcpy(ch[q],ch[np],sizeof(ch[q]));
	pre[q]=pre[np];len[q]=len[p]+1;
	pre[np]=pre[last]=q;
	for(;ch[p][c]==np;p=pre[p])
		ch[p][c]=q;
}

Dinic

int dis[N],head[N],Next[E],adj[E],k=1,S,T;
int que[N],now[N];
long long cap[E];
void addedge(int u,int v,long long w)
{
	Next[++k]=head[u],head[u]=k,adj[k]=v,cap[k]=w;
	Next[++k]=head[v],head[v]=k,adj[k]=u,cap[k]=0;
}
bool BFS()
{
	int i,l,r;
	memset(dis,-1,sizeof(dis));
	que[l=r=1]=S;
	dis[S]=0;
	while(l<=r)
	{
		for(i=head[que[l]];i;i=Next[i])
			if(dis[adj[i]]==-1&&cap[i])
			{
				dis[adj[i]]=dis[que[l]]+1;
				que[++r]=adj[i];
			}
		++l;
	}
	return (dis[T]!=-1);
}
long long DFS(int i,long long f)
{
	if(i==T)
		return f;
	long long ans=0;
	int j;
	for(j=now[i];j;now[i]=j=Next[j])
	{
		long long tmp;
		if(dis[adj[j]]==dis[i]+1&&cap[j]&&(tmp=DFS(adj[j],min(cap[j],f))))
		{
			cap[j]-=tmp;
			cap[j^1]+=tmp;
			f-=tmp;
			ans+=tmp;
			if(!f) break;
		}
	}
	return ans;
}
long long Dinic()
{
	long long ans=0;
	while(BFS())
	{
		memcpy(now,head,sizeof(head));
		ans+=DFS(S,inf);
	}
	return ans;
}

SPFA费用流

int head[N],Next[E],adj[E],cap[E],cost[E];
int q[N*20],pre[N],vis[N],now[N];
int flow,S,T,num=1;
long long d[N];
void addedge(int u,int v,int f,int w)
{
	Next[++num]=head[u],head[u]=num,adj[num]=v,cap[num]=f,cost[num]=w;
	Next[++num]=head[v],head[v]=num,adj[num]=u,cap[num]=0,cost[num]=-w;
}
bool SPFA()
{
	int i,l,r;
	q[l=r=1]=S;
	vis[S]=1;
	memset(d,0x7f,sizeof(d));
	d[S]=0;
	while(l<=r)
	{
		for(i=head[q[l]];i;i=Next[i])
			if(cap[i]&&d[adj[i]]>d[q[l]]+cost[i])
			{
				d[adj[i]]=d[q[l]]+cost[i];
				if(!vis[adj[i]])
				{
					vis[adj[i]]=1;
					q[++r]=adj[i];
				}
			}
		vis[q[l]]=0;
		++l;
	}
	return (d[T]<1<<30);
}
bool dfs(int i,int f)
{
	if(i==T)
	{
		flow=f;
		return true;
	}
	if(vis[i])
		return false;
	int j;
	vis[i]=1;
	for(j=now[i];j;now[i]=j=Next[j])
		if(cap[j]&&d[adj[j]]==d[i]+cost[j]&&dfs(adj[j],min(f,cap[j])))
		{
			cap[j]-=flow;
			cap[j^1]+=flow;
			vis[i]=0;
			return true;
		}
	vis[i]=0;
	return false;
}
long long ans,ansf;
void minmax()
{
	while(SPFA())
	{
		memcpy(now,head,sizeof(now));
		while(dfs(S,inf))
			ans+=flow*d[T],ansf+=flow;
	}
}

多项式合集

namespace poly{
    int R[N*4];
    long long qpow(long long a,long long b){
        long long ans=1;
        while(b){
            if(b&1)
                ans=ans*a%M;
            a=a*a%M;
            b>>=1;
        }
        return ans;
    }
    long long wn[N*4],iwn[N*4],inv[N*4],fac[N*4],ifac[N*4];
    void init(int E){
        int i;
        iwn[E/2]=wn[E/2]=1;
        long long s1=qpow(3,(M-1)/E);
        long long s2=qpow(s1,M-2);
        for(i=E/2+1;i<E;++i){
            wn[i]=wn[i-1]*s1%M;
            iwn[i]=iwn[i-1]*s2%M;
        }
        for(i=E/2-1;i;--i){
            wn[i]=wn[i<<1];
            iwn[i]=iwn[i<<1];
        }
        ifac[0]=fac[0]=inv[1]=1;
        for(i=2;i<E;++i)
            inv[i]=inv[M%i]*(M-M/i)%M;
        for(i=1;i<E;++i){
            ifac[i]=inv[i]*ifac[i-1]%M;
            fac[i]=fac[i-1]*i%M;
        }
    }
    unsigned long long ccc[N*4];
    void NTT(long long *f,int lim,int op){
        int i,j,k;
        for(i=0;i<lim;++i){
            R[i]=(R[i>>1]>>1)|(i&1?lim>>1:0);
            if(R[i]<i)
                swap(f[R[i]],f[i]);
        }
        for(i=0;i<lim;++i)
            ccc[i]=(f[i]%M+M)%M;
        for(i=1;i<lim;i<<=1)
            for(j=0;j<lim;j+=(i<<1))
                for(k=j;k<j+i;++k){
                    long long w=(op==1?wn[k-j+i]:iwn[k-j+i]);
                    unsigned long long p=ccc[k+i]*w%M;
                    ccc[k+i]=ccc[k]+M-p;
                    ccc[k]+=p;
                }
        for(i=0;i<lim;++i)
            f[i]=ccc[i]%M;
        if(op==-1){
            long long inv=qpow(lim,M-2);
            for(i=0;i<lim;++i)
                f[i]=f[i]*inv%M;
        }
    }
    long long ta[N*4],tb[N*4];
    void mult(long long *a,int n,long long *b,int m,long long *c){
        int lim=1;
        while(lim<n+m)
            lim<<=1;
        copy(a,a+n,ta);
        copy(b,b+m,tb);
        for(int i=n;i<lim;++i)
            ta[i]=0;
        for(int i=m;i<lim;++i)
            tb[i]=0;
        NTT(ta,lim,1);
        NTT(tb,lim,1);
        for(int i=0;i<lim;++i)
            ta[i]=ta[i]*tb[i]%M;
        NTT(ta,lim,-1);
        copy(ta,ta+lim,c);
    }
    long long tmp[N*4],tans[N*4];
    void Getinv(long long *a,long long *ans,int lim){
        ans[0]=qpow(a[0],M-2);
        for(int i=1;i<lim;i<<=1){
            for(int j=i;j<(i<<2);++j)
                ans[j]=tans[j]=tmp[j]=0;
            for(int j=0;j<(i<<1);++j)
                tmp[j]=a[j];
            for(int j=0;j<i;++j)
                tans[j]=ans[j];
            NTT(tmp,i<<2,1);
            NTT(tans,i<<2,1);
            for(int j=0;j<(i<<2);++j)
                tmp[j]=tmp[j]*tans[j]%M*tans[j]%M;
            NTT(tmp,i<<2,-1);
            for(int j=0;j<(i<<1);++j)
                ans[j]=(2*ans[j]-tmp[j])%M;
        }
    }
    long long tinv[N*4];
    void Getln(long long *a,long long *ans,int n){
        for(int i=0;i<n-1;++i)
            ans[i]=a[i+1]*(i+1)%M;
        Getinv(a,tinv,n);
        mult(ans,n-1,tinv,n,ans);
        for(int i=n;i>=1;--i)
            ans[i]=ans[i-1]*inv[i]%M;
        ans[0]=0;
    }
    long long tln[N*4];
    void Getexp(long long *a,long long *ans,int n){
        ans[0]=1;
        for(int i=1;i<n;i<<=1){
            for(int j=i;j<(i<<1);++j)
                ans[j]=0;
            Getln(ans,tln,i<<1);
            for(int j=0;j<(i<<1);++j)
                tln[j]=-tln[j]+a[j];
            ++tln[0];
            mult(ans,i,tln,i<<1,ans);
        }
    }
    void Getroot(long long *a,long long *ans,int n){
        ans[0]=1;
        for(int i=1;i<n;i<<=1){
            fill(ans+i,ans+(i<<1),0);
            Getinv(ans,tinv,i<<1);
            mult(tinv,i<<1,a,i<<1,tinv);
            for(int j=0;j<(i<<1);++j)
                ans[j]=(ans[j]+tinv[j])*inv[2]%M;
        }
    }
    long long ttln[N*4];
    void Getpow(long long *a,long long *ans,int n,int m){
        Getln(a,ttln,m);
        for(int i=0;i<m;++i)
            ttln[i]=ttln[i]*n%M;
        Getexp(ttln,ans,m);
    }
};

杜教筛

const int N=3000000;
int prime[N/5],vis[N+5],phi[N+5],mu[N+5],k,i;
long long sm[N+5],sp[N+5];
long long smu[1000005],sphi[1000005];
long long Smu(int m)
{
	if(m<=N)
		return sm[m];
	int t=n/m;
	if(smu[t])
		return smu[t];
	smu[t]=1;
	for(int l=2,r;l<=m;l=r+1)
	{
		r=m/(m/l);
		smu[t]-=(r-l+1)*Smu(m/l);
	}
	return smu[t];
}
long long Sphi(int m)
{
	if(m<=N)
		return sp[m];
	int t=n/m;
	if(sphi[t])
		return sphi[t];
	sphi[t]=1ll*m*(m+1)/2;
	for(int l=2,r;l<=m;l=r+1)
	{
		r=m/(m/l);
		sphi[t]-=(r-l+1)*Sphi(m/l);
	}
	return sphi[t];
}
void xxs()
{
	int i,j;
	mu[1]=phi[1]=1;
	for(i=2;i<=N;++i)
	{
		if(!vis[i])
		{
			mu[i]=-1;
			phi[i]=i-1;
			vis[i]=1;
			prime[++k]=i;
		}
		for(j=1;i*prime[j]<=N;++j)
		{
			int t=i*prime[j];
			vis[t]=1;
			if(i%prime[j]==0)
			{
				mu[t]=0;
				phi[t]=phi[i]*prime[j];
				break;
			}
			else
			{
				mu[t]=mu[i]*mu[prime[j]];
				phi[t]=phi[i]*phi[prime[j]];
			}
		}
	}
	for(i=1;i<=N;++i)
	{
		sm[i]=sm[i-1]+mu[i];
		sp[i]=sp[i-1]+phi[i];
	}
}

Dijkstra

int k,head[N],Next[N*2],adj[N*2],leng[N*2],vis[N],n;
long long d[N];
struct str{
	long long d;
	int x;
};
bool operator <(str a,str b)
{
	return a.d>b.d;
}
priority_queue<str> q;
void Push(int u,int v,int w)
{
	Next[++k]=head[u];
	head[u]=k;
	adj[k]=v;
	leng[k]=w;
}
void Dij(int S)
{
	int i,j;
	for(i=1;i<=n;++i)
	{
		d[i]=1000000000000000000ll;
		vis[i]=0;
	}
	d[S]=0;
	q.push((str){0,S});
	while(!q.empty())
	{
		str x=q.top();
		q.pop();
		if(vis[x.x])
			continue;
		vis[x.x]=1;
		for(j=head[x.x];j;j=Next[j])
			if(d[adj[j]]>d[x.x]+leng[j])
			{
				d[adj[j]]=d[x.x]+leng[j];
				q.push((str){d[adj[j]],adj[j]});
			}
	}
}

主席树(以单点加为例)

int ch[N*20][2],tree[N*20],k;
void pushup(int i)
{
	tree[i]=tree[ch[i][0]]+tree[ch[i][1]];
}
void modify(int i,int l,int r,int x,int y)
{
	if(l==r)
	{
		++k;
		tree[k]=tree[i]+y;
		return;
	}
	int mid=l+r>>1;
	int u=++k;
	if(mid>=x)
	{
		ch[u][1]=ch[i][1];
		ch[u][0]=k+1;
		modify(ch[i][0],l,mid,x,y);
		pushup(u);
	}
	else
	{
		ch[u][0]=ch[i][0];
		ch[u][1]=k+1;
		modify(ch[i][1],mid+1,r,x,y);
		pushup(u);
	}
}
void Build(int i,int l,int r)
{
	if(l==r)
		return;
	int mid=l+r>>1;
	ch[i][0]=++k;
	Build(ch[i][0],l,mid);
	ch[i][1]=++k;
	Build(ch[i][1],mid+1,r);
}
int Query(int i,int l,int r,int x)
{
	if(l==r)
		return tree[i];
	int mid=l+r>>1;
	if(mid>=x)
		return Query(ch[i][0],l,mid,x);
	else
		return Query(ch[i][1],mid+1,r,x);
}

部分几何模板

struct str{
	int x,y;
}q[500005];
struct pt{
	double x,y;
};
pt operator -(pt a,pt b)
{
	return (pt){a.x-b.x,a.y-b.y};
}
pt operator +(pt a,pt b)
{
	return (pt){a.x+b.x,a.y+b.y};
}
double area(pt x,pt y)
{
	return fabs((x.x*y.y-x.y*y.x)/2);
}
double dot(pt x,pt y)
{
	return x.x*y.x+x.y*y.y;
}
double dis(pt x,pt y)
{
	return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));
}
double D(pt x,pt y,pt a)//点到线段距离
{
	if(dot(a-x,y-x)<0||dot(a-y,x-y)<0)
		return min(dis(x,a),dis(y,a));
	return area(x-a,y-a)*2/dis(x,y);
}

批量生成输出模板

#include<bits/stdc++.h>
using namespace std;
int i;
string nm="string";
void OP(int x)
{
	string c="sub"+to_string(x)+"_"+nm+to_string(++id[x]);
	system(("copy "+c+".in "+nm+".in").data());
	system((nm+".exe").data());
	system((string("copy ")+nm+".out "+c+".out").data());
}
int main()
{
	for(i=1;i<=10;++i)
		OP(1);
}

咕咕咕……

posted @ 2020-03-17 18:18  夜螢光  阅读(224)  评论(0编辑  收藏  举报